Sed：从行中提取正则表达式模式

Question

I have an input stream of many lines which look like this:

path/to/file:             example: 'extract_me.proto'
path/to/other-file:             example: 'me_too.proto'
path/to/something/else:             example: 'and_me_2.proto'
...

I'd like to just extract the *.proto filenames from these lines, and I have tried:

[INPUT] | sed 's/^.*\([a-zA-Z0-9_]+\.proto\).*$/\1/'

I know that part of my problem is that .* is greedy and I'm going to get things like e.proto and o.proto and 2.proto , but I can't even get that far... it just outputs with the same lines as the input. Any help would be greatly appreciated.

Answer 1

I find it helpful to use extended regex for this purpose ( -r ) in which case you need not escape your brackets.

sed -r 's/^.*[^a-zA-Z0-9_]([a-zA-Z0-9_]+\.proto).*$/\1/'

The addition of [^a-zA-Z0-9_] forces the .* to not be greedy.

Answer 2

Since you tag your command with linux , I'll assume you have GNU grep. Pick one of

grep -oP '\w+\.proto' file
grep -o "[^']+\\.proto" file

Answer 3

one way to do it:

sed 's/^.*[^a-zA-Z0-9_]\([a-zA-Z0-9_]\+\.proto\).*$/\1/'

• escaped the + char
• put a negation before the alphanum+underscore to delimit the leading chars

another way: use single quote delimitation, after all it's here for that:

sed "s/^.*'\([a-zA-Z0-9_]\+\.proto\)'.*\$/\1/" 

Answer 4

Use this sed :

sed "s/^.*'\([a-zA-Z0-9_]\+\.proto\).*$/\1/"

+ - Extended-RegEx. So, you need to escape to get special meaning. The preceding item will be matched one or more times.

Another way:

sed "s/^.*'\([^']\+\.proto\)'.*$/\1/"

Answer 5

使用GNU sed：

sed -E "s/.*'([^']+)'$/\1/"